Question 6.7: Obtain the separated differential equations for potential di...

This example, like Example 6.5, further illustrates the method of separation of variables. Since the region is free of charge, we need to solve Laplace’s equation in cylindrical coordinates; that is

\nabla^2 V=\frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho\frac{\partial V}{\partial \rho}\right)+\frac{1}{\rho^{2}}\frac{\partial^{2} V}{\partial \phi^{2}}+\frac{\partial^{2} V}{\partial z^{2}}=0                    (6.7.1)

We let

V(\rho,\phi,z)=R(\rho)\Phi(\phi)Z(z)                                                    (6.7.2)

where R,  \Phi, and Z are, respectively, functions of \rho,  \phi, and z. Substituting eq. (6.7.2) into eq. (6.7.1) gives

\frac{\Phi Z}{\rho}\frac{d}{d\rho}\left(\frac{\rho dR}{d\rho}\right)+\frac{RZ}{\rho^{2}}\frac{d^{2}\Phi}{d\phi^{2}}+R\Phi\frac{d^{2}Z}{dz^{2}}=0                         (6.7.3)

We divide through by R\Phi Z to obtain

\frac{1}{\rho R}\frac{d}{d\rho}\left(\frac{\rho dR}{d\rho}\right)+\frac{1}{\rho^{2}\Phi}\frac{d^{2}\Phi}{d\phi^{2}}=-\frac{1}{Z}\frac{d^{2}Z}{dz^{2}}                           (6.7.4)

The right-hand side of this equation is solely a function of z, whereas the left-hand side does not depend on z. For the two sides to be equal, they must be constant; that is

\frac{1}{\rho R}\frac{d}{d\rho}\left(\frac{\rho dR}{d\rho}\right)+\frac{1}{\rho^{2}\Phi}\frac{d^{2}\Phi}{d\phi^{2}}=-\frac{1}{Z}\frac{d^{2}Z}{dz^{2}}=-\lambda^{2}                      (6.7.5)

where -\lambda^{2} is a separation constant. Equation (6.7.5) can be separated into two parts:

\frac{1}{Z}\frac{d^{2}Z}{dz^{2}}=\lambda^{2}                                                                           (6.7.6)

or

Z^{"}-\lambda^{2}Z=0                                                                          (6.7.7)

and

\frac{\rho}{R}\frac{d}{d\rho}\left(\frac{\rho dR}{d\rho}\right)+\lambda^{2}\rho^{2}+\frac{1}{\Phi}\frac{d^{2}\Phi}{d\phi^{2}}=0                                               (6.7.8)

Equation (6.7.8) can be written as

\frac{\rho^{2}}{R}\frac{d^{2}R}{d\rho^{2}}+\frac{\rho}{R}\frac{dR}{d\rho}+\lambda^{2}\rho^{2}=-\frac{1}{\Phi}\frac{d^{2}\Phi}{d\phi^{2}}=\mu^{2}                         (6.7.9)

where \mu^{2} is another separation constant. Equation (6.7.9) is separated as

\Phi^{"}+\mu^{2}\Phi=0                                                 (6.7.10)

and

\rho^{2}R^{"}+\rho R^{'}+(\rho^{2}\lambda^{2}-\mu^{2})R=0                         (6.7.11)

Equations (6.7.7), (6.7.10), and (6.7.11) are the required separated differential equations. Equation (6.7.7) has a solution similar to the solution obtained in Case 2 of Example 6.5; that is

Z(z)=c_{1}\cosh\lambda z+C_{2}\sinh \lambda z                         (6.7.12)

The solution to eq. (6.7.10) is similar to the solution obtained in Case 3 of Example 6.5; that is

\Phi(\phi)=c_{3}\cos\mu\phi+c_{4}\sin\mu\phi                         (6.7.13)

Equation (6.7.11) is known as the Bessel differential equation and its solution is beyond the scope of this text.^{1}